Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $x = \dfrac{a - 1}{a^2 - 4a + 3} \div \dfrac{5a^2 + 30a}{a^3 - 2a^2 - 3a} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $x = \dfrac{a - 1}{a^2 - 4a + 3} \times \dfrac{a^3 - 2a^2 - 3a}{5a^2 + 30a} $ First factor out any common factors. $x = \dfrac{a - 1}{a^2 - 4a + 3} \times \dfrac{a(a^2 - 2a - 3)}{5a(a + 6)} $ Then factor the quadratic expressions. $x = \dfrac {a - 1} {(a - 3)(a - 1)} \times \dfrac {a(a - 3)(a + 1)} {5a(a + 6)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac {(a - 1) \times a(a - 3)(a + 1) } { (a - 3)(a - 1) \times 5a(a + 6)} $ $x = \dfrac {a(a - 3)(a + 1)(a - 1)} {5a(a - 3)(a - 1)(a + 6)} $ Notice that $(a - 3)$ and $(a - 1)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac {a\cancel{(a - 3)}(a + 1)(a - 1)} {5a\cancel{(a - 3)}(a - 1)(a + 6)} $ We are dividing by $a - 3$ , so $a - 3 \neq 0$ Therefore, $a \neq 3$ $x = \dfrac {a\cancel{(a - 3)}(a + 1)\cancel{(a - 1)}} {5a\cancel{(a - 3)}\cancel{(a - 1)}(a + 6)} $ We are dividing by $a - 1$ , so $a - 1 \neq 0$ Therefore, $a \neq 1$ $x = \dfrac {a(a + 1)} {5a(a + 6)} $ $ x = \dfrac{a + 1}{5(a + 6)}; a \neq 3; a \neq 1 $